Space-time interval Lorentz co/invariant

1. The problem statement, all variables and given/known data
Hi guys!
I must show by brute force that ##\Delta s^2=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2## is invariant under Lorentz transforms.

Are you sure you can't orient your axes so that the x-axis is parallel to the boost?

I see, thank you very much.
Due to the complexity of the expression for a boost in any direction, I guess I can orient the axes so that it results to be in the x-direction. I will consider this case only.

At first glance I get ##\Delta s'^2=\Delta t^2 (-c^2\gamma^2+v^2)+\gamma ^2 \left ( \Delta x^2+v^2 \Delta t^2 - \frac{v^2\Delta x^2}{c^2} \right )##.
I guess I'll have to redo the math.
Edit:Yeah I forgot 2 terms. I reach now ##\Delta s'^2=\Delta t^2[v^2+\gamma(v^2-c^2)]+\Delta x^2+\Delta y^2+\Delta z^2##. Almost the desired result but apperently I still have some job to do :)

Edit 2:This reduces to ##\Delta s'^2=\Delta t^2 (v^2-c^2)+\Delta \vec x##. I'm almost there. But I would reach the result only if v=0... which is not correct.

Trace back to see how you got the first term of ##v^2## in the first brackets. I don't think that should be there.

I've tracked it down (already on my own), at ##\Delta x'=\gamma (\Delta x - v \Delta t)##. When I must take the square of this quantity, there's a term ##\gamma ^2 v^2\Delta t^2##. I couldn't cancel this term out though I must re-recheck the algebra.

Edit: Nevermind, I made a mistake. I now reach the desired result! Thank you guys.